∫ In trapezoidal ABCD, AD∨BC

Also made (auxiliary line making)

∴∠B=∠DEC

∫AB = DC = AD = 1/2BC。

∴DE=DC=EC

△DCE is equilateral △

∴∠B=∠DEC=60

2.e is the midpoint, because ADBE is an equilateral diamond.

because

AB=DC=AD= 1/2BC

therefore

namely

AB=BE=EC=CD=AD=DE= 1/2BC

3. It is a diamond for the following reasons:

Proof: ∵ABCD is a diamond.

∴AB∥CD,BC∥AD

And ∵P is any point on the diagonal AC (point P does not coincide with point A and point C), PE//BC in E intersects with AB, and PF//CD in F intersects with AD.

∴AB∥PF

PE∨AD

(that is to say, AE∨PF.

EP∑AF)

∴AEPF is a parallelogram.

Also, ABCD is a diamond.

∴BC=CD

∴PE=PF

∴AEPF is a diamond.

Is point P the intersection of AC and BD?

Let's be clear about this.

The position of point P is not clear, and the area is different.